Register

ISRO 2008- ECE adjoint Matrix

• edited by
1,498 views

1 Answer

2 votes
2 votes
| adj A| = |A|^(n-1) And here A= adj(A)

So |A| = |A|^2 So either |A| =0 or 1 also you can put all values and check or solve by using |A| = 0 or 1 and answer will be B) = 3
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
3 answers
1
sh!va asked Mar 4, 2017
398 views
ISRO 2006-ECE Matrix determinant
The determinant of is: a) 0 b) -1 c) 1 d) 2
The determinant of is:a) 0b) -1c) 1d) 2
User Avatar
0 votes
0 votes
2 answers
2
sh!va asked Mar 4, 2017
565 views
ISRO 2006-ECE Eigen Values
The Eigen values of are: a) 6, 1 b) 6, -1 c) -6, 1 d) -6,-1
The Eigen values of are:a) 6, 1b) 6, -1c) -6, 1d) -6,-1
User Avatar
0 votes
0 votes
1 answer
3
sh!va asked Mar 4, 2017
1,850 views
ISRO 2006-ECE Functions
$x^3$ + $x$ sin $x$ is a) Constant function b) Odd function c) Even function d) Periodic function
$x^3$ + $x$ sin $x$ isa) Constant functionb) Odd functionc) Even functiond) Periodic function
User Avatar
0 votes
0 votes
1 answer
4
sh!va asked Mar 4, 2017
379 views
ISRO 2006-ECE Concurrent Lines
The value of k for which the lines 2x + y -1 = 0,4x + 3y - 3 = 0 and 3x + ky - 2 = 0. are concurrent is a) -2 b) 3 c) 2 d) -3
The value of k for which the lines 2x + y -1 = 0,4x + 3y - 3 = 0 and 3x + ky - 2 = 0. are concurrent isa) -2b) 3c) 2d) -3
User Avatar
Link Copied!